Binary vector expected value
Some lines above the author says 


  By the law of large numbers we can also assume that the total input to the node has a Gaussian distribution 


hence we can assume $X \sim \mathcal{N}(0,1)$ with the $X$ domain being continuous 

Then he says the input vector is assumed to be binary which changes the domain from continuous to discrete so we can discretize it assuming $-1 \le X \le 1$ is mapped into zero and $X&lt; -1 $ and $X &gt; 1$ are mapped to 1 

Finally according to the 68-95-99.7 Rule we can compute 

$$ E(X^2) = P(X=1) \cdot 1^2 + P(X=0) \cdot 0^2 = 0.32 $$

Finally probably the author rounds this up to $\frac{1}{3} \simeq 0.33$